Continued Fractions and Hyperelliptic Curves

I recently read a charming little paper: Quasi-elliptic integrals and periodic continued fractions, by van der Poorten and Tran. Most of us who have taken a number theory course of some kind learned how to solve Pell’s equation: x^2 - D y^2 =1 where D is a nonsquare positive integer. The usual method is to compute the continued fraction
\displaystyle{\sqrt{D} = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{\cdots}}}}.
One then defines the convergents of \sqrt{D} by
\displaystyle{x_0/y_0 = a_0}
\displaystyle{x_1/y_1 = a_0 + \frac{1}{a_1}}
\displaystyle{x_2/y_2 = a_0 + \frac{1}{a_1+\frac{1}{a_2}}} etcetera.

Then x_i^2 - D y_i^2 tends to be very small and, if you compute long enough, for some i you will have x_i^2 - D y_i^2=1.

What van der Poorten and Tran do is to ask what happens if D is not an integer, but a polynomial D(t) = t^{2g+2} + d_{2g+1} t^{2g+1} + \cdots + d_1 t + d_0. Before I get into details, I want to tell you about something gorgeous that I won’t explain at all. Using the methods in their paper, van der Poorten and Trap can discover identities like
\displaystyle{ \int \frac{3 x dx}{\sqrt{x^4+2x}} = \log \left( x^3+1+x \sqrt{x^4+2x} \right)}.
Isn’t that pretty?

It turns out that the continued fraction algorithm for \sqrt{D(t)} is actually much prettier than for integers. Everything should be understood in terms of the curve C cut out by y^2 = D(t). This is a curve of genus g, with two points at infinity. (One of these points is the limit of (t, \sqrt{D(t)}) and the other is the limit of (t, -\sqrt{D(t)}).) I’ll call these two points \infty_{+} and \infty_{-}. The theory is controlled by the line bundles \mathcal{O}(k \infty_+ + \ell \infty_-). In particular, there are nontrivial solutions to x(t)^2 - D(t) y(t)^2 =1 if and only if the continued fraction is periodic, if and only if \mathcal{O}(k \infty_+) = \mathcal{O}(k \infty_-) for some a >0.

Below the fold, I’ll explain what is meant by the continued fraction algorithm for an algebraic function, and tell you some of the other nice results from the paper.

Given any power series Y(t) = Y_k t^k + Y_{k-1} t^{k-1} + \cdots in t^{-1}, we define the continued fraction of Y(t).

Define [Y(t)] := \sum_{i=0}^k Y_i t^i. Set a_0 = [Y] and define F_1 by Y = a_0 + 1/F_1. Then set a_1 = [F_1] and F_1 = a_1 + 1/F_2. Continuing in this way, we get a sequence a_i of polynomials, a sequence F_i of power series, and a continued fraction
\displaystyle{Y = a_0 + \frac{1}{a_1+\frac{1}{a_2+\frac{1}{\cdots}}}}.
We can also define the convergents x_i/y_i as before; they do converge to Y in the sense that each ratio x_i/y_i agrees with Y to a higher order than the ratio does.

In particular, suppose that D(t) is a polynomial of the form t^{2g+2} + d_{2g+1} t^{2g+1} + \cdots + d_1 t + d_0.
Then \sqrt{D(t)} is a power series in t^{-1}:
\displaystyle{\sqrt{D(t)}} = t^{g+1} + (1/2) d_{2g+1} t^g + \cdots.
So we can define the continued fraction of \sqrt{D(t)}.
We keep the notations a_i(t), F_i(t), x_i(t) and y_i(t) from above.

I’ll explain just one key idea from the paper. Let’s think about the zeroes and poles of F_i(t). Since a_i(t) is a polynomial in t, its only poles are at \infty_{\pm}, and it has a pole of the same order at both \infty‘s. So, other than \infty_{\pm}, the function F_i(t) - a_i(t) has the same poles as F_i. Then F_{i+1} = 1/(F_i - a_i) has zeroes at the poles of F_i - a_i.

That’s what happens away from \infty_{\pm}. Suppose that F_i(t) has a pole of order p > 0 at \infty_{+}, and a zero of order q > 0 at \infty_{-}. Then a_i(t) has a pole of order p at both \infty‘s. The difference F_i(t) - a_i(t) has a zero of order \geq 1 at \infty_{+} and a pole of order p at \infty_{-}. So F_{i+1} has a pole of order \geq 1 at \infty_{+} and a zero of order p at \infty_{-}.

Summing up the last two paragraphs, let the poles of F_i be P + p \infty_{+} and let the zeroes be Q + q \infty_{-}. Then the poles of F_{i+1} are R+r \infty_{+}, for some R and some r \geq 1 and the zeroes are P + p \infty_{-}. (Here P, Q and R are supported away from \infty_{\pm}.) In other words, there is a sequence of positive integers p_i and a sequence of divisors P_i such that the poles of F_i are P_i + p_i \infty_{+} while the zeroes are P_{i-1} + p_{i-1} \infty_{-}.

Note that p_i is the degree of a_i. Note also that P_i + p_i \infty_{+} \equiv P_{i-1} + p_{i-i} \infty_{-} in the Picard group, so
P_i \equiv P_0 + p_0 \infty_{-} + \sum p_j (\infty_{-} - \infty_{+}) - p_{i} \infty_{+}.

It’s not too hard to work out what happens if the coefficients of D(t) are chosen generically. The first a_0 has degree g+1 and all the other p_i are 1. A bit of effort checks that P_1 has degree g (exercise!), so all of P_i have degree g and, in fact, P_i \equiv (g+i) \infty_{-} - i \infty_{+} in the Picard group. You may remember that a generic divisor of degree g has a unique effective representative in PIcard; P_i is that unique representative. So, we have just found an explicit way to write down an arithmetic progression in Pic^g(C), where C is a hyperelliptic curve.

Of course, the fun comes in the nongeneric case. In that case, the p_i can skip around. It’s really fun when \infty_{+} - \infty_{-} is torsion in the Picard group or, in other words, when there is a unit x(t) + y(t) \sqrt{D(t)} in the coordinate ring of C. Then, eventually, the sequence in Picard will repeat. It turns out, when this happens, the corresponding approximation x_i(t)/y_i(t) gives your unit!

There are plenty of other ideas in the paper. What is the analogue of the result that the x_i/y_i are the best approximations to \sqrt{D}? The F_i are all of the form (A_i + \sqrt{D})/B_i: how do we relate the polynomials A_i and B_i to the divisors P_i? And how did I come up with that integral above? All this and more, so read the paper!

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4 thoughts on “Continued Fractions and Hyperelliptic Curves”

  1. I don’t understand what is meant by

    [Y(t)] :=\sum_{i=0}^k Y_k t_k

    Are some of those k’s perhaps supposed to be i’s or something?

  3. I’m pretty sure Kenny’s right, so I corrected the post moving the subscript to a superscript. David if I did this wrong please change it back.

  4. Thanks for the fix, but you missed the other typo: the summation variable is i, not k. Fixed now.

    Kenny’s English summary is completely correct.

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